 One, we have Olivier Wittenberg who will be talking around the inverse gamma problem second lecture. Thank you. Can you hear me well? Yeah. OK. So before I begin, let me remind you that there's a wine and talk graduate students event tonight at 6 PM in the dining tent. OK. So let me remind you what we did yesterday. So I talked about the Nutter problem, which was the rationality of this variety, the quotient of An by G acting by permutation on the coordinates. And I explained that it's a stronger problem than Grunwald's problem, which is itself a stronger problem than the inverse gamma problem. And I explained that Nutter's problem has a negative answer in general. I gave a counter example, which I explained through a counter example to Grunwald's problem for z mod 8z and q. So I should just add that for cyclic groups and over q, the answer to Nutter's problem is by now completely understood. So this combines two things. First of all, theoretical advances, in particular one of the early works of Hendrik Landstra, and extensive computer calculations. I mean, the theoretical advances tell you that for a cyclic group acting on a fine space over q, the Nutter's problem has a positive answer, if and only if. Some arithmetic property of some cyclotomic field holds true, but then you need to do something more. I mean, it's not easy to check. So then there are also ingredients coming from number theories and height estimates. And finally, as I said, extensive computer calculations. So I encourage you to take a look at the notes where this is explained. So I'm not going to talk about this today. And also in the notes, I explained that there are also counter examples to Nutter's problem for over the complex numbers, of course, for non-Abelian groups then. OK, so what are we going to do today? So OK, Nutter's problem has a negative answer in general. Remember why we were interested in a positive answer? It was in order to apply Hilbert's irreducibility theorem to the projection from an to this quotient. If the base is rational, then you can find lots of points such that the fiber is irreducible, is an irreducible, sorry, connected g-tosser. And then you solve the inverse general problem. So if it's not rational, you can do it like this. So let's today take another point of view. Let's look for a torsor to which we could apply Hilbert's irreducibility theorem, but maybe not this one. Let's take the simplest one for which the base is rational. So let's look at torsos over p1, open subsets of p1. So that's the so-called regular inverse general problem, which you can ask over any field for any finite group g, finite group. So the question is, does there exist a connected g-tosser over an open subset of p1? And let's even ask for a bit more, geometrically connected. So does there exist a geometrically connected g-tosser? So y to x, where x is an open of p1. So just to make things clearer, let me draw a picture. So this is x, sorry, this is p1. And x is you remove a few points from p1. And y is going to be a curve mapping to p1, mapping to x, like this. So the g-tosser will be over the points where there's no ramification. So you remove these points. And there's an action of g on the curve upstairs, which acts fiber-wise and which acts simply transitively on the fibers. That's why. So in particular, the extension of function fields is Galois with group g. And so what does geometrically connected mean? Well, it's not only a connected curve, but even if you extend the scalars from k to an algebraic closure, it remains connected. That's what it means. So if you can do this, and k is, say, q, then you apply Hilbert's irreducibility theorem and you immediately solve the inverse Galois problem for g over q. So it's, again, a stronger version. But contrary to Nutter's problem, this might well always have a positive answer for all fields and all finite groups. This is probably what's hoped. At least no counter example is known. OK, and what's good with this question, better than with the inverse Galois problem itself, is that it has this geometric flavor which connects it to topology in a very close way. And topology is much easier than number theory. So here's the theorem which really makes the connection. So it's the so-called Riemann's existence theorem. So you take x and open in p1. So this is over the complex numbers. x in p1 over the complex numbers and open. You remove finitely many points. Then you can look at the curves covering x, which are finite et al. So all the morphisms, pi from y to x, finite et al morphisms. So here I'm not requiring that these have an action of g and so on. So in this situation, really the picture is this. And finite et al means there's no ramification. So it extends to some curve which maps to p1 with some ramification. But over x, there's no ramification. So when you have this, you can look at the complex points. So it's topological space. And you get a topological covering. So finite topological coverings of the complex points of x. And the way you do this, again, is start from this and you just look at the complex points of y. So remember, topological covering is a topological space mapping to x of c. And locally on the base, it's trivial. It's a product. And finite means that the fibers are finite. So you have this thing. And the theorem is this is an equivalence of categories. In other words, you have a dictionary between this algebraic world here and this purely topological world. And that's very useful. So this has two consequences, the corollary. So first, well, let's take a connected g-torso over x. So here, again, x is an open in p1 of the complex numbers. So let's take a connected g-torso. Then so, OK, we have a connected g-torso. g is just the group of automorphisms of y over x. What this dictionary tells us is that we can compute this group of automorphism in the purely topological setting. It's also just the group of automorphism of the topological covering. So that's what topologists call deck transformations. So this is a very nice consequence because it tells you if you want to compute this Galois group, you just have to compute something purely topological. And the second consequence is that if you want to construct a finite detail covering of x, an algebraic curve like this, it's enough to construct a finite topological covering. And then you get an algebraic structure on the top space for free. So we are interested in g-torso. So let me state it for g-torso. Sorry, I will go there. So two. So from this dictionary, you deduce that a connected g-torso over x is the same thing as what? So what is a connected g-torso? Well, it's a finite detail cover like this, which is connected. And which in addition has, so connected means it's not a distant union of two such things. And in addition, there's an action of g on it, which is simply transitive on the fibers. Well, the dictionary tells you it's the same thing as a finite topological covering, which is connected. It's not a distant union of two things. With an action of g, which is simply transitive on the fibers. But this is something we understand from topology. When you have a nice topological space like this, there's this notion of universal covering. You fix the base point. You have the universal cover. And there is a Galois theory of topological coverings. So the universal cover has a group of automorphisms, which is the fundamental group of the base. And all connected covers are factors through it. Sorry, I mean, are sub-covers of this. It's just the same thing as in Galois theory for fields. And so there's a notion of Galois, finite topological coverings. So that's a connected topological covering whose group of automorphism is big enough, is equal to the degree of the covering. And so you can realize them always as intermediate covers of the universal covering. So that means you can realize the group as a quotient of the fundamental group. So I'm sorry, I'm going a bit fast, but this is pure topology. It's written in all topological textbooks. And it tells you that by this dictionary, what you get is exactly the surjections from the fundamental group of the base. So you pick a base point, small x, to g, up to conjugation by g, by an element of g. So again, let me just draw what I just said. You have this space and there's this universal covering. Here the automorphism group is pi 1. And if you have your connected g torsor here, it will correspond to a sub covering. OK, so you get g as a quotient of pi 1. But you have to choose a way to factor it and that's why you get only this up to conjugation. OK, but now this is very nice because the fundamental group is very simple. We're looking at an open in p1. So let's say x is p1 minus some points which I call b1 to br. So here they would be the branch points here under the ramification points, b1, br, and so on. Well, everyone knows that the fundamental group of this, if you take the complex points, the sphere minus r points, this is a free group on r minus 1 generators. And to be a bit more canonical, this really has a presentation with r generators, gamma 1 to gamma r, and one relation. The product of them is 1. So let me draw another picture. So here's the complex plane and there's the points at infinity. We have the points b1, b2, br. And we have our base point, which we have fixed, x. So the gamma i's are the loops based at x, which go around each bi like this. So these are generators and it's clear that on the picture the composition is trivial. We're in the sphere, there's a point at infinity. OK, so now let's plug this into that. So a morphism like this is the same as giving the images of the gamma i's. So that's the same as giving a tuple of elements g1, gr in g, such that the product is 1, and such that the g i's generate the group, because I want this to be onto. OK, up to conjugation, sorry, up to conjugation. So this is in gr, so you have r elements of g, and g acts by simultaneous conjugation on all of the components. So this is something very explicit. So in particular, this gives a direct answer to the regular inverse problem over the complex numbers. So the thing is, this set is non-empty if r is large enough. That's obvious. And so by this dictionary, we do get a connected g cover, g torso over x if r is large enough. So corollary, positive answer to the regular inverse problem over c for any g. So again, the thing is, you pick out large enough. You choose the g i's. This gives you a topological covering. And by Riemann, you can algebra it and get your cover. OK, this dictionary is useful also for non-Galois cover. So let me mention something a bit more precise. So let's pick finite total cover like this. Again, x is my open in p1 over c. And let's assume y is connected. OK, then so from topology, we know that we get a topological covering by looking at the complex points. But more than that, we get a monodromy action. So the fundamental group of the base acts on the fiber. So let's pick x. So pi 1 of the base acts on the fiber, pi 1 of x. So if you number these, I mean, so sorry. So this is the monodromy action. So if this is x, this is the fiber, you start with a loop base that x, and you lift it starting from one point. This will give you a path leading to another point. So let's number the elements of the fiber. We get a map to the symmetric group, sn. Say n is the cardinality of the fiber. And so what we know from topology is that if you take, so this is maybe not Galois, but you can take a Galois closure. And then the automorphism group of the Galois closure is the image of this monodromy map. OK, the image. Yeah. So g, Galois group of a Galois closure, let me say of the extension of function fields, right? So sorry, what the corollary says is introduce g, the Galois group of a Galois closure of this function field extension. Then you can view it in a topological way. Namely, it's the image of this monodromy map. So it's a certain quotient of the fundamental group of the base. And it's actually even better than that. You can describe what the gamma i's here do as permutations of the fiber. So namely, gamma i acts. So sorry, gamma i acts on the fiber as a product of cycles. And the lengths of the cycles are given by the ramification indices above bi as a product of cycles whose lengths are the ramification indices above bi. OK, so here what do I mean exactly? So we started with something which is only over x, a finite total curve over x. Well, there's nothing above bi. So there's a way to compactify y to a curve mapping to p1. So we have this picture, y to x. We can compactify it. There's a unique smooth projective curve containing y as an open subset, and it paths to p1. And now this is not unromified anymore. So here you have ramification above bi, and that's what I mean here. OK, so the first part is really just translation using the dictionary. And here to know that gamma i acts in this way, it's a purely local computation. You just have to know how a ramified morphism of curves looks like locally on the complex points. So I mean, OK, this is a standard thing. And this is very useful in practice. It allows you to compute the Galois group of a Galois closure when you give yourself a cover of curves like this. So for example, let me not enter into any detail here, but for example, if you have a curve which covers p1 like this, and the ramification is as simple as it can be. That is, of course, there will be some ramification, but it will only be ramification index 2. And you don't have two ramification points in the same fiber, so that's really the generic situation. Then what this tells you is that if you take the Galois closure, so that's going to be another curve, which now is going to be Galois over p1, then the group here is generated by elements. So it's a subgroup of the symmetric group, and it's generated by elements which act as transpositions. Here it is just above any bi. One ramification point of index 2, and that's it. And then it's a standard lemma in group theory, such a subgroup of the symmetric group, which is generated by transpositions and which acts transitively on the fiber, that's the connectedness assumption, must be SN, the whole of SN. So this shows that the Galois closure, sorry, that G equals SN if the ramification is as simple as possible. So it's on purpose that I'm not going into any detail here. I just want to show an application of this, and you will see more in the exercises. So am I going to separate? So do you mean two ramification points above the same point of p1? So this group here is generated by the gamma i's, and each gamma i will be a transposition. So if you have two ramification points like this, you will have two gamma i's, each of which will be a transposition. Does that answer your question? I'm sorry, that's OK. So let's go back to the inverse Galois problem. So OK, we get a positive answer over C here. Now of course, we're interested at things over Q. So the whole question now is how to get down from C to Q. And as far as I know, all answers to the regular inverse Galois problem over any field first go through this over C, and then you try to descend. So there is one method, one technique, which is at the same time very specific and incredibly effective. It's called the rigidity method. So I will try to explain this now. So the rigidity method. So let me state the theorem. So this is due to many people, Belly, Frid, Matzat, Thompson. And I will state it just in the basic case. There are many variants on this, but already this basic case works very well. So you fix an integer R. It's going to be the number of branch points. You fix a finite group G. And let's assume that the group has trivial center. Let's in addition fix a collection of conjugacy classes in G, C1 to CR conjugacy classes. OK, and then the theorem says that, yes, we can realize G as a regular inverse problem has a positive answer for G over Q under some assumptions on the CIs. So the regular inverse problem for G over Q has a positive answer if, and I'm going to state the conditions. So the conditions are at the same time a bit technical and completely elementary if. So first of all, each CI is so-called rational. So this means that for any G in CI and any integer n prime to the order of G, if you look at G to the power n, it's conjugated to G in G, in capital G. G and GM are conjugates. OK, and the second condition is that the collection C1 to CR is so-called rigid. And this means the following. So you see this set here. I'm going to almost rewrite it. And I'm going to assume in addition that each GI belongs to the conjugacy class CI. So you look at the tuples G1, GR. So each GI is in CI. And I assume that the product is one and that they generate the group. OK, and the condition is that if I look at this modular, the action of G by conjugation, I get just one element. So in other words, if I don't quotient, I'm going to say G acts transitively on this set. G acts by simultaneous conjugation transitively on this set. So I'm assuming that the set is non-empty on this set, which is non-empty. OK, that's part of the assumption. OK? So as I said, it's a very down-to-earth statement. And what's maybe surprising is that it's extremely useful. So there are many groups which have been shown to be Galois groups of a cube because we could apply this theorem to them. So in particular, the assumptions are known to hold for at least 10 of the sporadic simple groups exactly in this form with even a very small value of r equals 3. And for many other classes of groups. So for example, you can apply this to PSL2 of fp for infinitely many values of p, and so on and so forth. So it's not easy to apply this theorem. You have to be able to find these conjugacy classes satisfying these assumptions. When they exist, sometimes they don't exist. And if you try to do it, I mean you quickly run into computational challenges. People have used computers to look for such things. And in fact, as soon as r gets a bit large, I mean even 5 or 6, very quickly it becomes infeasible. Also, there's some series group theory going into verification of this. But OK, it can be done. And another thing I should say is that when this theorem applies, in principle, it is possible to write down explicitly an equation for the g toss for y that this theorem would produce. So I give some references in the notes. I encourage you to look at this. And what I will try to do today is not explain how to apply this, but rather explain where these conditions come from and why this is true. OK, so let's fix once and for all a group g and the integer r. So let's use a terminology, so a g cover over k. So this is going to be a geometrically connected g torso, just a shorthand, a geometrically connected g torso over an open subset of p1, open x in p1 over k. OK, so that's what we're looking for, a g cover over q. So it turns out that it's very useful to look at the set of all g covers. So there's a modular space for this. So let me first introduce just one thing. Whenever I have such a thing, I will denote by capital B the set of branch points. So b will be p1 minus x. OK, so it turns out that there is a modular space for these things. So that means there's a sum variety whose points, whose k points, are the g covers over k for any k. So if you fix the degree of b. So that's a theorem due to Fridt and Thurkline. So sorry, and here comes one of the assumptions. The center of g should be trivial. If the center of g is trivial, there is a variety, curly h over q, that parameterizes g covers whose branch point, whose branch locus has degree r. So you have r branch points. So g covers with degree of b equals r. So here parameterizes is in a loose sense, but I really mean very precisely that for any field k of character 6-0, the set of k points is the set of isomorphism classes of g covers over k with r branch points. g covers over k with degree of b equals r. That's right. That's right. Yes, thank you. So I fix r, but I don't fix b. OK, so that's something you have to take as a black box. I mean, this thing exists. And so it's a very standard thing when you solve modular problems in general, that you're never going to find such a thing if the objects you're trying to classify have non-trivial automorphisms. And this is the meaning of this assumption here. You can check. It's easy. That's the automorphisms of g cover. That's the center of g. And moreover, sorry, not only is there such a variety, but in addition, and there is a finite et al map from this script curly h to curly u. And curly u is the variety which parametrizes the b's. That parametrizes the b's. So this is very easy to construct. It's an open in PR, actually, the b's. I mean, b is given by the zero locus of some polynomial, which is well defined up to scalars. And the coefficients of this polynomial give you a point in PR. So this is an open in PR. OK, and the map is what? Well, you start with a g cover and you associate to it the underlying b, a g cover y to x. And you map it to b. So the statement is, this is actually finite et al. It's a very nice topological cover. So it already says a lot about the geometry of h. And OK, so now that we have this manualized space, all we want to do is to find a rational point on h. That's all we want to do. So goal h of q, non-empty. OK, to reach this goal, we need to understand the geometry a little better. And for this, we will need to associate with the g cover. Not only its branch locus, but also the classes g i, the conjugacy classes of the g i's that we saw earlier, the images of the gamma i's, the loops around b i. Yeah. The way I'm telling the story today, they don't need to ramify. But in old papers, they are supposed to actually ramify. In all the literature on the topic, the assumption is that the b i's ramify. That's in the literature. But today, I'm not doing this. Today, for simplicity, I'm not assuming that the b i's ramify. So exactly, that's why I said a g torso over an open x. So x is part of the data. Yeah, but it's a minor point. But thank you for the question. This I don't know. I mean, OK. If you know that a certain group satisfies the assumptions of the rigidity theorem, can you get some extensions of this group which also satisfy the assumptions? I don't know. I'm sorry, I'm not an expert on this. I don't know. OK, so remember, we had this associated with our g cover. We had this map, a suggestion well-defined up to conjugacy from the topological fundamental group. And here, we have the generators gamma i, which are mapped to g i. And so we want to take this into account. The problem is this definition is purely topological. So it's never going to work in this arithmetic setting over q. So we need to find an algebraic analog. So here, I'm going to tell what it is. So algebraic version of this. So there's an analogy. So analogy. So what is gamma i? You're looking really at b i in the complex situation. And you have a small disk around b i. You remove b i and you look at a loop, which goes around b i counterclockwise. So you're looking at the fundamental group of a disk minus the center, the unit disk in the complex plane with some base point. So this is, well, this is z generated by a counterclockwise loop around 0. OK, in the algebraic setting, the analogous thing is going to be the absolute Galois group of q bar double parenthesis t. That's what you get when you localize at t equals 0 from the function field of p1. And what is this Galois group? Well, this is the field of Puezer series. So you just need to extract nth roots of t. So you get these cyclic extensions. And the Galois group, when you extract an nth root of t, is not it's cyclic, but it's not z more than z. It's mu n. So this is what one usually calls z hat twist 1, which is the inverse limit of all of the mu n's. So mu n of q bar. So it's a procyclic group, but there's not a canonical generator as we had here. So that's why things become a bit more tricky. And so what was GI? GI was the image of the generator here by the map that you get from pi1 of the disk around bi minus bi to pi1 of the whole x and then to g. Well, in the same way, here, because we're looking here, our g is the Galois group of the function field extension. And the function field of p1 maps to q bar double parenthesis t. Well, I mean, you can localize it and complete it at bi. And so this inclusion of fields gives you naturally a morphism between the absolute Galois groups in the other direction. And that's how you get, in a canonical way, a morphism from this group to g. So really, sorry, I continue the analogy. So GI, which is really a map from the conjugacy class of GI, which is a map from z to g up to conjugacy. So this is well-defined only up to conjugacy, remember. The analogy is going to be a map from z hat of 1, a continuous map from this to g, up to conjugacy. OK, so that's something here. So let me not be too precise with the definitions. I'm telling you, this is the analogy. So from a g cover, you naturally get at each bi a homomorphism like this in a canonical way up to conjugation. So the point is that now we are going to look at things over q. And if you look at things over q, you want to keep track of Galois actions, of the action of Galois of q bar over q. And here we do have a non-trivial Galois action. It acts on this through this dichotomy character. OK, so now let's use this. So what we want to do now is we have this map. So sorry, I will write it here. That's the end of the analogy. We have this map from h to u. So this parameterizes g covers. And starting from this parameterizes the b's. Starting from the g cover, you associate with the underlying b. What we want to do is to associate with not just the b, but also the conjugacy classes of the GIs above each bi. So we're going to have to consider a variety here, which is going to parametrize this data. And then we will get a factorization like this. So what is b going to be? So first, I'm telling you we have this thing. This is a finite set. And there's an action of Galois of q bar over q over it. So you can view it as the q bar points of some variety of dimension 0 defined over q. There is c, a variety of dimension 0 over q, such that the q bar points of c are exactly this homomorphisms from z hat twist 1 to g up to conjugation. And this is an equality of sets with the action of Galois of q bar over q equivalent. So this is what parametrizes our GIs. And so b is going to parametrizes the pairs consisting of a b and a map from b to c pairs b. So b is something of degree r in p1 and a map from b to c. And I mean this really in the same precise sense that I alluded to. So this is a variety over q. And for any field of characteristic 0, the k points of this are the 0 dimensional subvarieties of p1 over k of degree r, and maps of between varieties over k from this b to this c. So in the same way that u was very easy to construct as an open PR, v is easy to construct by hand too. So let me not insist on this. So there's some v. And now what I'm telling you is that because we did everything canonically, there is a factorization here. And so you can send a g cover here, y to x to the pair consisting of the branch locus b and the conjugacy classes of the gis. So this is something I will write it as clear of g i, even though there's no g i here. But it's the thing which lives here and which is canonical. So conjugacy classes of all of the gis. And so you project just by forgetting the conjugacy classes. So I'm telling you this is the geometry. And now what the rigidity method does for you is the following. So let me go here. So remember, we want to construct a rational point on h. So for this, we have to construct a rational point on v first. So let's start with a rational point on u. And let's start with the simplest possible rational point. We fix branch points b1 to br. And I will choose them to be rational points. So I choose points in rational points of p1, where I was distinct. So then the capital B, which is the collection of these points, gives me a rational point of u. And now you can wonder, what are the rational points of v above this one? Rational points of v above this one, b. Well, they are exactly the maps from b to c. But b is just a collection of rational points. So they are just collections of rational points of c. So that's really c of q cross c of q r times. But now what is c of q? It's a homomorphism like this, from that hat to 1 to g, which is up to conjugation, and which is Galway invariant up to conjugation. And now if you think just a little bit about it, you realize this is exactly a rational conjugacy class. So rational in the sense that I stated in the theorem, rational conjugacy classes. So where does that come from? So really the point is that the Galway group acts here through the cyclotomy character. So you really want, I mean, you don't think to be invariant up to conjugation when you act by the cyclotomy character. But on q, the cyclotomy character is surjective. So you want to have something which is invariant under conjugation whenever you raise to any power prime to the order of the element. This is really a translation. So I'll leave it to you. And so OK, we have a rational point on v. But we want a rational point on h. And that's where this method is really unreasonable. To construct a rational point on h, you would need to understand the geometry of h. But the theory, h could be something very complicated. What turns out is that the hypothesis of the theorem, rigidity, implies something extremely strong, namely, imply that this map is in fact an isomorphism in the neighborhood of our point. And so in a sense, we get the rational point on h for free. So the fact, so if you fix a point of v above b, so that's fixing rational conjugacy classes, so h to u, to v, sorry, is an isomorphism locally above our point b, c1, cr, if the collection is rigid. And how do you see this? Well, to see that it's an isomorphism, locally, remember, this is a finite isomorphism, it's enough to look at the complex points and to see that on the complex points, it's locally an isomorphism. But on the complex points, we understand everything very well. So remember that the fiber on the complex points of the map from h to u above b, well, what is it? It's the set of isomorphism classes of g covers over the complement of b. And we saw as a consequence of Riemann's existence theorem that such things are exactly the tuples g1, gr of elements of g, which generate g and whose product is one up to conjugation, g1, gr in gr, such that the product is one and g1, gr generate g up to conjugation. And then we're interested in the fiber not from h to u, but from h to v. So what is the map in these terms from h to v? Well, starting from such a thing, you associate to it the conjugacy class of g1, the conjugacy class of gr. And so the fiber of h of c to v of c above b comma c1, cr is, well, it's the subset of this where you impose in addition the condition that each di belongs to ci, subset where for any i gi belongs to ci. And so you see the rigidity is exactly telling you that this fiber has cardinality 1, singleton if c1, cr is rigid. So you get the local isomorphism of the complex, so also over q. And so your rational point lives to rational point of h. So that's a bit extreme. Hopefully there should be other ways to produce rational points of h. But it seems very difficult to understand the geometry of h. So even such a simple question as understanding the irreducible components of h is difficult. But this is something concrete which you can understand with a computer, because this is a finite data covering, even over the complex numbers. This is a finite data covering. So it's described by the action of the fundamental group of u on a finite set, the fiber. And the fundamental group of u is completely explicit. It's a braid group. So you have this problem, you have a braid group, and you try to understand the orbits of the braid group. And again, people have been doing this. But as soon as r gets large enough, it becomes very difficult even for modern computers. OK, a lot of time. I'll stop here for today. Thank you for your attention. Thank you.